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{\bf Question}

Let $\{S_m\}$ be a sequence of sets.

Let $\ds H_n=S_n\cap\left(\bigcup_{m<n}S_m\right)^C$

Show that $\ds H_m\cap H_n=\phi \hspace{0.2in}$ for $m\not=n$, and
that $\ds \bigcup_{n=1}^{\infty}H_n=\bigcup_{n=1}^{\infty}S_n$



\vspace{0.25in}

{\bf Answer}

Suppose without loss of generality $m<n$

$\ds H_n=S_n\cap\bigcap_{t<n}(S_t)^C\subseteq(S_m)^C$

$\ds H_m=S_m\cap(\bigcup_{t<m}(S_t))^C\subseteq S_m$

Hence $H_n\cap H_m=\phi$

${}$

Since $\ds H_n\subseteq S_n, \hspace{0.2in} \bigcup_{n=1}^\infty
H_n\subseteq\bigcup_{n=1}^\infty S_n$

Now suppose $\ds x\epsilon\bigcup_{n=1}^\infty S_n$.  Let $r$ be
the smallest integer such that $x\epsilon S_r$, then
$x\not{\epsilon}S_m$, for $m<r$.

Therefore $\ds x\epsilon S_r\cap(\bigcup_{m<r}S_m)^C=H_r$

Therefore $\ds x\epsilon\bigcup_{r=1}^\infty H_r$.

Hence the result.




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